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NEET CHEMISTRYAlcohols, Phenols and EthersMedium

Question

Compound A (C8H10OC_8H_{10}O) is found to react with NaOI (produced by reacting Y with NaOH) and yield a yellow precipitate with a characteristic smell. A and Y are respectively?

A

2-Phenylethanol and Iodine

B

Acetophenone and Iodine

C

1-Phenylethanol and Chlorine

D

1-Phenylethanol and Iodine

Step-by-Step Solution

  1. Identify the Reaction: The formation of a yellow precipitate with a characteristic smell upon reaction with sodium hypoiodite (NaOI) is the specific confirmatory test for Iodoform (CHI3CHI_3).
  2. Identify Reagent Y: NaOI is prepared in situ by the reaction of Iodine (I2I_2) with Sodium Hydroxide (NaOH). Therefore, Y is Iodine. (I2+2NaOHNaOI+NaI+H2OI_2 + 2NaOH \rightarrow NaOI + NaI + H_2O)
  3. Identify Compound A: The iodoform test is positive for compounds containing the methyl ketone group (CH3C=OCH_3-C=O) or the methyl carbinol group (CH3CH(OH)CH_3-CH(OH)-) linked to a carbon or hydrogen. The molecular formula is given as C8H10OC_8H_{10}O. 1-Phenylethanol (C6H5CH(OH)CH3C_6H_5-CH(OH)-CH_3) fits this formula (C8H10OC_8H_{10}O) and contains the required secondary alcohol group (CH3CH(OH)CH_3-CH(OH)-) attached to a phenyl ring. Upon oxidation by NaOI, it converts to acetophenone and then undergoes the haloform reaction to yield iodoform (CHI3CHI_3) and sodium benzoate. Note: 2-Phenylethanol (C6H5CH2CH2OHC_6H_5-CH_2-CH_2-OH) does not have the required grouping. Acetophenone (C6H5COCH3C_6H_5COCH_3) reacts but its formula is C8H8OC_8H_8O, not C8H10OC_8H_{10}O.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Alcohols, Phenols and Ethers. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYAlcohols, Phenols and Etherscompoundproducedreactingyellowprecipitate

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